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Need some help with math

  1. Feb 19, 2005 #1
    A sequence [tex]t_{1}[/tex],[tex]t_{2}[/tex],...[tex]t_{n},...[/tex] is defined as follows:

    [tex]t_{1}=14[/tex]
    [tex]t_{k}=24-5t_{k-1}[/tex] for each k <= 2

    For every positive integer [tex]n[/tex], [tex]t_{n}[/tex] can be expressed as [tex]t_{n}=p*q^n+r[/tex] where p, q and r are constants. The value of p, q and r is?

    I found out [tex]t2=-46[/tex] and [tex]t3=254[/tex] and wrote out

    [tex]14=p*q+r\\[/tex]
    [tex]-46=p*q^2+r\\[/tex]
    [tex]254=p*q^3+r[/tex]

    I don't know what to do from here... is what I did so far even correct? TiA.
     
  2. jcsd
  3. Feb 19, 2005 #2

    arildno

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    First of all, the recurrence relation ought to be:
    [tex]t_{k}=24-5t_{k-1},k>=2[/tex]
    Now, for arbitrary k, you have been given:
    [tex]t_{k}=pq^{k}+r,t_{k-1}=pq^{k-1}+r[/tex]
    Put this into the recurrence relation, and rearrange a bit:
    [tex](q+5)pq^{k-1}=24-6r[/tex]
    This equality must hold for EVERY k>=2. What does that suggest to you?
     
  4. Feb 19, 2005 #3
    I don't really get what's happening...
     
  5. Feb 19, 2005 #4

    HallsofIvy

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    Well, the first thing that happened is that you had a "misprint". You had written "for each k<= 2" when you clearly meant "for each k>= 2". That was what arildno was talking about.

    THEN his point was that, if you know tn= p qn+ r, you can put that into the recursion formula. Replacing n by n-1 gives tn-1= p qn-1+ r so the recursion formula says p qn+ r= 24-5(p qn-1+ r). That equation is true for all n and has 3 unknown numbers in it so you can evaluate it for 3 values of n (2, 3, 4 will do) to get 3 equations for those 3 unknowns.
    That's exactly what you did so, yes, what you have done so far is correct.

    Now, to solve those equations:

    An obvious thing to do is subtract the first equation from the second so that you get
    60= pq- pq2=pq(1-q).

    Subtract the second equation from the third to get 300= pq3- pq2= pq2(1-q).

    Now, from the first equation, p= 60/(q(1-q)) so the second equation is
    300= 60 (q2(1-q))/(q(1-q)). That is, 300= 60 q or q= 5.
    Now work back to solve for p and r.
     
  6. Feb 19, 2005 #5
    I got it... thanks for your help!

    ...but now I have another question that I'm stuck on. I can't even begin to do anything because I have no idea how to find out the answer... so some hints or something would be appreciated.

    A circle that has a radius of 3 is tangent to both the positive x-axis and the positive y-axis. Another circle with a radius of 1 is tangent to the bigger circle and to the positive x-axis. There is a line tangent to both circles. Where does this line cross the y-axis?

    I had to change the question around a bit... there's a picture associated that makes it eaiser to imagine but its not that hard to draw out.... and the line isn't the x-axis!
     
  7. Feb 20, 2005 #6

    xanthym

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    Refer to the following diagram:
    http://img110.exs.cx/img110/7287/tangentcircles4lh.jpg
    From the RED Right-Triangle construct, determine the coordinates of Points A, B, C, and D. The equation of Line AC through points A and C can then be found, and thus the equation of Line ED through Point E and Normal to Line AC at Point D can be determined. The Y-intercept of Line ED is the solution to your problem.
     
  8. Feb 21, 2005 #7
    Thanks for the help... I have another (simple) question: How do I evaluate (3^2000+3^-2000)^2-(3^2000-3^-2000)^2
     
  9. Feb 22, 2005 #8

    Curious3141

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    Better to put different questions into different threads.

    For the last one, use [tex]a^2 - b^2 = (a+b)(a-b)[/tex] and see what you get.
     
  10. Feb 22, 2005 #9
    Ill do that in the future... I just dont know how to simplify (do I even simplify?) something like 3^2000+3^2000...

    this is the solution... but I don't understand where 2.3 came from and how it becomes 4.3, etc...

    [​IMG]
     
  11. Feb 22, 2005 #10

    xanthym

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    {2.32000} should be interpreted {2*32000}
    {4.30} should be interpreted {4*30}


    ~~
     
    Last edited: Feb 22, 2005
  12. Feb 22, 2005 #11
    wicked... it makes sense now =)
     
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